topic 5: probability distributions

Topic 5: Probability Distributions

Achievement Standard 90646

Solve Probability Distribution Models to solve straightforward

problems

4 CreditsExternally Assessed

NuLake Pages 278 322

NORMAL DISTRIBUTION

PART 2

Lesson 3: Making a continuity correction

• Go over 1 of the final 2 qs from HW (combined events). NuLake p303.

• How to calculate normal distribution probabilities using your Graphics Calculator.

To practice using GC: Do Sigma (NEW – photocopy): p358 – Ex. 17.01 (Q3 only). Write qs on board as a quiz.

• Continuity corrections – how and when to make them.

Work: Fill in handout on cont. corr., then NuLake p309: Q42-46. Finish for HW.

*Don’t do Q47.

Using your Graphics Calc. for Standard Normal problems

• MENU, STAT, DIST,NORM; then there are three options: * Npd – you will not have to use this option * Ncd – for calculating probabilities * InvN – for inverse problems

• NB: On your graphics calculator shaded areas are from -∞ to the point.

– To enter -∞ you type – EXP 99.

– To enter +∞ you type EXP 99.






175 =178 184

=5

E.g. 1:

If =178, =5, P(175 X 184) = ?

MENU, STAT, DIST, NORM, Ncd

lower: 175, upper: 184, σ: 5, μ: 178






175 =178 184

=5

E.g. 1:

If =178, =5, P(175 X 184) = 0.61067


lower: 175, upper: 184, σ: 5, μ: 178





175 =178 184

=5

E.g.1:

If =178, =5, P(175 X 184) = 0.61067


lower: 175, upper: 184, σ: 5, μ: 178E.g.2:

If =30, =3.5, P(X 31) = ?


lower: -EXP99, upper: 31, σ: 3.5, μ: 30

31 =30

=3.5





175 =178 184

=5

E.g.1:

If =178, =5, P(175 X 184) = 0.61067


lower: 175, upper: 184, σ: 5, μ: 178E.g.2:

If =30, =3.5, P(X 31) = 0.61245



31 =30

=3.5





175 =178 184

=5

E.g.1:

If =178, =5, P(175 X 184) = 0.61067


lower: 175, upper: 184, σ: 5, μ: 178E.g.2:

If =30, =3.5, P(X 31) = 0.61245



31 =30

=3.5

10 minutesDo Sigma (NEW) p358. Ex. 17.01• Q3 on the board as a quiz.

Making a Continuity Correction

(USE THE HANDOUT)

17.02AHeights of Year 13 males in NZ are normally distributed with mean 174 cm and standard deviation 6 cm. If they are measured to the nearest cm, calculate the probability that the height of a student is more than 165 cm.

174165

174165

Distribution when rounding heights (discrete histogram)

Actual curve for heights of students (continuous)

165 165.5164.5

For example the probability that a student was 165 cm tall would have to be represented by a column with base 164.5 to 165.5

174165

174165

165 165.5164.5 165 165.5164.5




Heights of Year 13 males in NZ are normally distributed with mean 174 cm and standard deviation 6 cm. If they are measured to the nearest cm, calculate the probability that the height of a student is more than 165 cm.

174165

174165

because any student with a height in this interval would be recorded as having a height of 165 cm.

165 165.5164.5 165 165.5164.5





174165

174165

165 165.5164.5

165 165.5164.5

P(X > 165) = P(X > ?) with continuity correction.

To find the cut-off point for continuity corrections, move up or down to the midpoint between two whole-numbers.

In this example the wording is ‘more than 165’, so move up to 165.5.


174165

174165

165 165.5164.5165 165.5164.5

P(X > 165) ≈ P(X > 165.5) with continuity correction.

To find the cut-off point for continuity corrections, move up or down to the midpoint between two whole-numbers.

In this example the wording is ‘more than 165’, so move up to 165.5.


≈ 0.9217 (4 sf)

Continuity CorrectionIf we use a Normal Distribution to approximate a variable that

is DISCRETE, we must make a Continuity Correction.

DISCRETE CONTINUOUS

Continuity CorrectionIf we use a Normal Distribution to approximate a variable that is

DISCRETE, we must make a Continuity Correction.

DISCRETE CONTINUOUS

P(X = 4)



DISCRETE CONTINUOUS

P(X = 4) P(3.5 < X < 4.5)



DISCRETE CONTINUOUS

P(X = 4) P(3.5 < X < 4.5)

P(X > 6)



DISCRETE CONTINUOUS

P(X = 4) P(3.5 < X < 4.5)

P(X > 6) P(X > 6.5)



DISCRETE CONTINUOUS

P(X = 4) P(3.5 < X < 4.5)

P(X > 6) P(X > 6.5)

P(X > 6)



DISCRETE CONTINUOUS

P(X = 4) P(3.5 < X < 4.5)

P(X > 6) P(X > 6.5)

P(X > 6) P(X > 5.5)



DISCRETE CONTINUOUS

P(X = 4) P(3.5 < X < 4.5)

P(X > 6) P(X > 6.5)

P(X > 6) P(X > 5.5)

P(X < 10)

P(X < 10)

P(8 < X < 12)

Do NuLake qs on “Continuity Corrections for a Normal Distribution”:Pg. 311-313: Q4246 (NOTE: Don’t do Q47).



DISCRETE CONTINUOUS

P(X = 4) P(3.5 < X < 4.5)

P(X > 6) P(X > 6.5)

P(X > 6) P(X > 5.5)

P(X < 10) P(X < 9.5)

P(X < 10)

P(8 < X < 12)




DISCRETE CONTINUOUS

P(X = 4) P(3.5 < X < 4.5)

P(X > 6) P(X > 6.5)

P(X > 6) P(X > 5.5)

P(X < 10) P(X < 9.5)

P(X < 10) P(X < 10.5)

P(8 < X < 12)




DISCRETE CONTINUOUS

P(X = 4) P(3.5 < X < 4.5)

P(X > 6) P(X > 6.5)

P(X > 6) P(X > 5.5)

P(X < 10) P(X < 9.5)

P(X < 10) P(X < 10.5)

P(8 < X < 12) P(7.5 < X < 12.5)


Lesson 4: Inverse normal problems where you are given the probability and asked to calculate the x-value.

Learning outcome:Calculate the x cut-off score based on

given probabilities, and a given mean and SD.

Work:1. Inverse calculations using standard normal.2. Inverse calculations – standardising Examples3. Do Sigma (new - photocopy): p366 – Ex. 17.03.

Inverse questions - the other way around

Where you’re told the probability and have to find the z-values.

Examples:(a) Find the value of z giving the area of 0.3770 between 0 and z.

0.377

0 zz = ?

P(0 < Z < z) = 0.377

What is z ?Answer (from tables): z = ?

P(0 < Z < z) = 0.377

What is z ?Answer (from tables): z = 1.16




0.377

0 zz = 1.16




0.377

0 zz = 1.16

(b)Find the value of z if the area to the right of z is only 0.05.

0.05

0.45

0 zz = ?

P(0 < Z < z) = 0.45





0.377

0 zz = 1.16

(b)Find the value of z if the area to the right of z is only 0.05.

0.05

0.45

0 zz = 1.645

Inverse problems where you’re given the probability, and , and asked to

find the value of X.

E.g. A normally distributed random variable has a mean of 24 & std. deviation of 4.7. What value has only 5% of the distribution above it?

i.e. P(X > xcut-off) = 0.05.

We’re told that =24 and =4.7. What is the value, xcut-off ?

z = x –

can be re-arranged to solve for x

x = + z

z = x –


x = + z


i.e. P(X > xcut-off) = 0.05.


z = x –


x = + z


i.e. P(X > xcut-off ) = 0.05.


0.05

0.45

0 z= 24 xcut-off = ?

z = x –


x = + z


i.e. P(X > xcut-off ) = 0.05.

We’re told that =24 and =4.7. What is the value, xcut-off ?First do using working (standardise it), then check with G.Calc.

First find the z cut-off like in the last example

= 4.7

P(0 < Z < z) = 0.45


0.05

0.45

0 z= 24

xcut-off = ?

z = x –


x = + z


i.e. P(X > xcut-off ) = 0.05.


zcut-off = 1.645

0.05

0.45

0 z= 24

xcut-off =

z = x –


x = + z


i.e. P(X > xcut-off ) = 0.05.


zcut-off = 1.645

0.05

0.45

0 z

STAT, DIST, NORM, InvNArea: _____ , σ: 4.7, μ: 24

Area: Enter total area to the LEFT of xcut-off .

= 24

z = x –


x = + z


i.e. P(X > xcut-off ) = 0.05.


zcut-off = 1.645

xcut-off =

0.05

0.45

0 z

STAT, DIST, NORM, InvNArea: 1-0.05 , σ: 4.7, μ: 24


= 24

z = x –


x = + z


i.e. P(X > xcut-off ) = 0.05.


zcut-off = 1.645

xcut-off =

0.05

0.45

0 z

STAT, DIST, NORM, InvNArea: 1-0.05 , σ: 4.7, μ: 24 = ____


= 24

z = x –


x = + z


i.e. P(X > xcut-off ) = 0.05.


zcut-off = 1.645

xcut-off =

0.05

0.45

0 z

STAT, DIST, NORM, InvNArea: 1-0.05 , σ: 4.7, μ: 24 = 0.95


= 24

z = x –


x = + z


i.e. P(X > xcut-off ) = 0.05.


zcut-off = 1.645

xcut-off =

0.05

0.45

0 z

STAT, DIST, NORM, InvNArea: 1-0.05 , σ: 4.7, μ: 24 = 0.95


= 24

z = x –


x = + z


i.e. P(X > xcut-off ) = 0.05.


Once you’ve copied down the e.g. & working:

Do Sigma (NEW version): p366 – Ex. 17.03 Complete for HW.

Extension (after you’ve finished this): NuLake p307 & 308

xcut-off = 31.73

zcut-off = 1.645

Lesson 5: Inverse normal problems where you must calculate the mean or

SD

• Calculate the mean if given the SD and the probability of X taking a certain domain of values.

• Calculate the SD if given the mean and the probability of X taking a certain domain of values.

Sigma (new - PHOTOCOPY): p369, Ex 17.04.

STARTER:

Question from what we did last lesson:

Inverse Normal: Calculating the x cut-off score.

0.02

0.48

Inverse normal question: A manufacturer of car tyres knows that her product has a mean life of 2.3 years with a standard deviation of 0.4 years. Assuming that the lifetime of a tyre is normally distributed what guarantee should she offer if she only wants to pay out on 2% of tyres produced.Solution:

Let X be a random variable representing the life of a tyre.X is normal with μ = 2.3 and σ = 0.4

We want an x value such that P(X xcut-off ) = 0.02.

gives a z value of -2.054 (see tables )

So her guarantee should run for 1.48 years. AnswerNote: In practice, what would be a sensible guarantee?

xcut-off = ?xcut-off = 1.48 yrs

02.0)( zZP

= -2.054 4.0

3.2 offcutx

z

STARTER QUESTION(from what we did last lesson)

0.02

0.48

Inverse normal question: A manufacturer of car tyres knows that her product has a mean life of 2.3 years with a standard deviation of 0.4 years. Assuming that the lifetime of a tyre is normally distributed what guarantee should she offer if she only wants to pay out on 2% of tyres produced.

Solution:Let X be a random variable representing the life of a tyre.X is normal with μ = 2.3 and σ = 0.4

We want an x value such that P(X xcut-off ) = 0.02.

gives a z value of -2.054 (see tables )

So her guarantee should run for 1.48 years. AnswerNote: In practice, what would be a sensible guarantee? Perhaps 17 months?

02.0)( zZP

= -2.054 4.0

3.2 offcutx

z

xcut-off = 1.48 yrs

E.g. 1:P(X < 903) = 0.657. The standard deviation is 17.3. Calculate the mean.

Calculate the value of z from the information P(Z < z ) = 0.657

Note: z must be above the mean as the probability is > 0.5

903

X

0.657

Inverse Normal Problems where you’re asked to calculate the MEAN or STANDARD DEVIATION

E.g. 2:X is a normally distributed random variable with mean of 45. The probability that X is less than 37 is 0.02.Estimate the standard deviation of X.

E.g. 2: X is a normally distributed random variable with mean of 45. The probability that X is less than 37 is 0.02.Estimate the standard deviation of X.

2.0537

Z

0

Standard normal distribution 37

X

45

0.02

0.02

Calculate z using your graphics calc.Use the Standard Normal Distribution,so use InvN and enter:Area :0.02 :1 :0

z = 2.0537

P( Z ) = 0.02

We want such that P( X 37 ) = 0.02

P( Z ) = 0.02

4537 8

So P( Z -2.0537) = 0.02

37

X

45

0.02

Calculate z using your graphics calc.Use the Standard Normal Distribution,so enter:Area :0.02 :1 :0

z = 2.0537

P( Z ) = 0.02

We want such that P( X 37 ) = 0.02

P( Z ) = 0.02

4537 8

P( Z -2.0537) = 0.02

So = -2.0537

8

= 3.895 (4 sf)

Re-arrange to solve for .= 8 2.0537

Do Sigma (new edition) - p369, Ex 17.04

Lessons 6 : Sums & differences of 2 or more normally-distributed variables.

Learning outcome:Calculate probabilities of outcomes that

involve sums or differences of 2 or more normally-distributed random variables.

Work:1. Notes & examples on sums & differences2. Spend 15 mins on Sigma (NEW version): Ex.

18.01 (p377)3. Notes & example on totals of n identical

independent random variables.4. Finish Sigma Ex. 18.01 (complete for HW)

Probabilities when variables are combined

1. X and Y have independent normal distributions with means 70 and 100 and standard deviations 5 and 12, respectively. If T = X + Y, calculate:(a) The mean of T.(b) The standard deviation of T.

(c) P(T < 180)

2. A large high school holds a cross-country race for both boys and girls on the same course. The times taken in minutes can be modelled by normal distributions, as given in the table.

If a boy and girl are both chosen at random, calculate the probability that the girl finishes before the boy.

Boys Girls

Mean 28 35

Standard Deviation 5 4

Do Sigma (new version): pg. 377 – Ex. 18.01

Let the total passenger load be T

E(T) = E(X1) + E(X2) + . . . + E(X25)Write the formula for the mean.

E(T) = 25×E(X) Since the 25 distributions are identical.

E(T) = 25

In general, for n items with identical distributions, the Expected Value of the distribution of the total is given by:

E(T) = n

Suppose there is a total of 25 people in a lift. Each person, if chosen at random, has a mean weight of 65 kg with standard deviation 7 kg.(a) Find the mean and standard deviation of the total passenger load.


E(T) = E(X1) + E(X2) + . . . + E(X25)

E(T) = 25×E(X) Since the 25 distributions are identical.

E(T) = 25

E(T) = n

E(T) = n

= 25 65

= 1625 kg

In general, for n items with identical distributions, the Expected Value of the distribution of the total is given by:

3.01


E(T) = n

= 25 65

= 1625 kg

Write the formula for the mean.

Substitute and calculate.

To find the std. deviation, first work through the VARIANCE.Let the total passenger load be T

Var(T) = Var(X1) + Var(X2) + . . . + Var(X25)

Var(T) = 25×Var(X)

T = )(25 XVarTo find the Standard Deviation, take the square root of the Variance.



Var(T) = Var(X1) + Var(X2) + . . . + Var(X25)

Var(T) = 25×Var(X)

)(25 XVarTo find the Standard Deviation, take the square root of the Variance.T =

In general, for n items with identical distributions, the Standard Deviation of the distribution of the total is given by:

T= )(XVarn

3.01

= 35 kg

Write the formula for the standard deviation.

Substitute and calculate.

T =

XnT2


T= )(XVarn

2725

)(25 XVar

4925

In general, for n items with identical distributions, the Standard Deviation of the distribution of the total is given by:

3.01Suppose there is a total of 25 people in a lift. Each person, if chosen at random, has a mean weight of 65 kg with standard deviation 7 kg.(a) Find the mean and standard deviation of the total passenger load.


E(T) = n= 25 65

= 35 kg

= 1625 kg

XT n 2 2725

4925

So the total weight of the passengers, T, is approximately normally distributed with of 1625 kg and of 35 kg.

3.03

(b) The lift is overloaded when the total passenger load exceeds 1700 kg. Calculate the probability that the lift is overloaded, assuming that the lift is carrying 25 passengers.

Strategy: Find the mean and standard deviation of T, the total load.

As the distribution of T is approximately normal, use this information to calculate the probability of overload.

P(T > 1700) = 1700 1625P 35Z Calculate the probability that T > 1700.

= 0.01606 (4sf)

(b) The lift is overloaded when the total passenger load exceeds 1700 kg. Calculate the probability that the lift is overloaded, assuming that the lift is carrying 25 passengers.

Continue through Sigma Ex. 18.01. Complete for HW

E(T) = n= 25 65

= 35 kg

= 1625 kg

XT n 2 2725

4925

So the total weight of the passengers, T, is approximately normally distributed with of 1625 kg and of 35 kg.

Lessons 7 : Linear combinations of normally-distributed variables.

Learning outcome:Calculate probabilities of outcomes that

involve a linear function of a random variable or a linear combination of 2 random variables.

Work:1. Notes on linear combinations (re-cap of expectation)2. Sigma (NEW) – Ex. 18.02 (pg. 380) – do 1st 2 qs.3. Handout – distinguishing between totals & a linear

functions.4. Finish Sigma Ex. 18.02 (complete for HW). To Q4

compulsory. Q5 on extension.

Linear Function of a Random Variable, X

aX + c, (e.g. taxi fares: hourly rate per km + fixed

cost)Its mean E(aX+c) = a × E(X) + c

Its variance Var(aX+c) = a2 ×

Var(X)Its std. deviation σaX+c = )(2 XVara

Linear Combination of 2 independent random variables, X & Y

Distribution of aX + bY where a & b are

constantsIts mean E(aX + bY) = a×E(X) +

b×E(Y)

Its variance Var(aX+c) = a2 ×

Var(X)Its std. deviation σaX+c = )(2 XVara




b×E(Y)Its variance Var(aX + bY) = a2Var(X) + b2Var(Y)

Its std. deviation σaX+bY = )()( 22 YVarbXVara






Do Sigma (NEW): pg. 380 – Ex. 18.02 - Q1 and 2.1. X has a normal distribution with mean 40 & standard dev. of 3.(a)Calculate the mean & standard deviation of W where W = 6X + 15.

(a)Calculate P(W>250)





Its std. deviation σaX+bY =

Do Sigma (NEW): pg. 380 – Ex. 18.02 - Q1 and 2.2. X has a normal distribution with mean 18 & SD of 2.4, and Y has a norm. distn. with mean 22 & SD of 1.5. W=3X + 5Y.(a)Calculate the mean and SD of W.

(a)Calculate P(W<160)

HANDOUT ON DISTINGUISHING BETWEEN

LINEAR FUNCTIONS AND TOTALS

Distinguishing between Linear Functions and Totals of Identically Distributed

Variables.

A telephone contractor installs cable from the street to the nearest jackpoint inside a house. The length of cable installed for each job in a particular new subdivision can be modelled by a normal distribution X with a mean of 12m and a standard deviation of 1.6m.

What is the difference between the following 2 questions?

Situation 1 - Linear Function Situation 2 - TOTAL of identically-distributed variables.

There is a charge of $5 per metre. What is the probability that the job exceeds $70?

What is the probability that a group of 5 jobs require a total of more than 70m?

Situation 1 - Linear Function Situation 2 - Total



Multiplying our variable (nbr metres) by a co-efficient ($5 per metre).So it’s a Linear Function of ONE variable.

Total of 5 different random variables. It just so happens they have the same distribution.






There is one random variable, X, and the cost of the job is 5X.

There are 5 random variables:

X1, X2, X3, X4, X5.

The length of cabling is the sum of all 5.

Each variable has a mean 12 & Std. Dev 1.6.








X1, X2, X3, X4, X5.



The Model is: C = aX.

E(C) = a × E(X) and

VAR(C) = a2 × VAR(X), so

σC = √ (a2 × σ2X)

The Model is: T = X1+X2+ X3+X4+ X5.

E(T) = E(X1)+E(X2)+E(X3)+E(X4)+E(X5).

VAR(T) = VAR(X1)+VAR(X2)+…+VAR(X5).

σT = √ (σ2X1+ σ2

X2+…+ σ2X5)








X1, X2, X3, X4, X5.






σC = √ (a2 × σ2X)


E(T) = E(X1)+E(X2)+E(X3)+E(X4)+E(X5).


σT = √ (σ2X1+ σ2

X2+…+ σ2X5)

E(C) = 5 × 12 = $60

σC = √ (52 × 1.62) = 8

E(T) = 12 + 12 + 12 + 12 + 12 = 60m.

σT = √ (1.62+ 1.62+1.62+ 1.62+1.62)

= 3.5777




Multiplying our variable (nbr metres) by a co-efficient ($5 per metre).

So it’s a Linear Function of ONE variable.



There are five random variables:

X1, X2, X3, X4, X5.






σC = √ (a2 × σ2X)


E(T) = E(X1)+E(X2)+E(X3)+E(X4)+E(X5).


σT = √ (σ2X1+ σ2

X2+…+ σ2X5)

Parameters:

E(C) = 5 × 12 = $60

σC = √ (52 × 1.62) = $8

Parameters:

E(T) = 5 × 12 = 60m.

σT = √ (5 × 1.62) = 3.5777m

With μC = E(C) = $60 and σC = $8,

we get P(C > 70) = 0.1056 (4sf)

With μT = E(T) = 60m and σT= 3.577709,

we get P(T > 70) = 0.002594 (4sf)

Probability that one job cost > $70 is 0.1056 (4SF)

Probability that total length required for 5 jobs exceeds 70m is 0.002394 (4SF)

Continue through Sigma Ex. 18.02. Complete for HW

topic 5: probability distributions

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